Assigning values to functions (Newbie)

This is a discussion on Assigning values to functions (Newbie) within the C Programming forums, part of the General Programming Boards category; I have just started learning C and I already have some issues.
Code:
#include <stdio.h>
int binom(int n, int k)
...

mean it will assign "n" value from "main" function to "binom" function? (Same applies to "k")

What you would say in programming-speak is: n and k are submitted as arguments to binom, and the result is assigned to bnm.

I know this is stupid question, and yes I googled it but I havent found a decent answer. Thanks.

If you are unsure what something does, get as many hints as you can and then start playing around to see what happens. If you have a theory about what you think is happening, try to prove and disprove it.

Finally: LEARN TO INDENT. STARTING NOW. There are a bunch of examples in C for various "indent styles" here:

Thanks!
One more question. I don't understand the logic behind this... In the "main" function we keep incrementing n and k at the same rate until n=9.
Shouldn't that keep the ratio of n and k equal (n==k) and therefore always return 1?
I know I got something wrong but I'm not sure what...

Thanks!
One more question. I don't understand the logic behind this... In the "main" function we keep incrementing n and k at the same rate until n=9.
Shouldn't that keep the ratio of n and k equal (n==k) and therefore always return 1?
I know I got something wrong but I'm not sure what...

If you would learn to indent your code properly you would actually be able to see what is happening...

So that means that after testing if n<10 is true the program solves the "binom" function and prints it, increments n, goes into a new line, and after all that it looks if k<=n and does the orders in its brackets? I'm sorry if you find it difficult to understand my stupid question regarding this simple issue.

EDIT: "See it now? the k loop is inside the n loop... for each value of n you call binom with every value of k." I think I speak english well but this statement confused me and hence my even more confusing question.

I got it but I haven't said what I meant right.
I thought that the value of j stays saved if we get back to the first loop, I didn't know that after we get back to the second loop initialization is made again.

No j is not saved because you don't go back to the outer loop until the inner one is finished and each time you re-enter the inner loop it's starting over... the numerical printout from my example should have demonstrated that very clearly.